Astronomy Problems: Please Look/help!

Sorry, that was wrong too. I wasn't cubing the bottom. So the new mass is:
Mass = 3.30343624 x 10^-23 kg.

That must be right. I have double-checked and looked over my work/calculations.
 

SpaceTiger

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astronomystudent said:
Sorry, that was wrong too. I wasn't cubing the bottom. So the new mass is:
Mass = 3.30343624 x 10^-23 kg.

That must be right. I have double-checked and looked over my work/calculations.
First, I would suggest reflecting on how much a kilogram actually is. If necessary, do a google search on the masses of various objects and see if you can find something as massive as your Kuiper Belt object. If you're not sure what a Kuiper Belt object is, you might also want to look that up. Second, you may want to consider writing out your calculation. I can only guess as to where you are going wrong.
 

tony873004

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You still have your diameter wrong. **Edit... Never mind, you got it right**. You'll need to fix that before you can get your volume and density.

Your formula for mass is wrong. Let me give you an example of how to extract M from that formula. I'll use the escape velocity formula to illustrate. (You don't need this formula for your question).

Ve = 1.4142 * sqrt (GM/r)

M is inside a root symbol. So you have to get it out. Square everything:

Ve^2 = 1.4142^2 * sqrt (GM/r)^2

The sqrt and the ^2 cancel out leaving you with:

Ve^2 = 1.4142^2 * GM/r

Cross multiply:

Ve^2 x r = 1.4142^2 x GM

Divide to isolate M:

M = (Ve^2 x r) / (1.4142^2 x G)

Use the same procedure to isolate M from your orbital period formula.
 
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SpaceTiger

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tony873004 said:
You still have your diameter wrong. You'll need to fix that before you can get your volume and density.
Are you sure about that? I'm getting the same result.


Your formula for mass is wrong.
Haw. Yeah, I just noticed that his/her formula for mass is inverted. His/her method for obtaining it from Kepler's Third Law was probably alright. I suspect he/she just made an algebra mistake.
 
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tony873004

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Oops. I didn't notice he recomputed from his original 20 km answer. 456 km is correct.
 
Isn't it 456 meters, not km? I am finishing the problem. It is due in an hour..
 

SpaceTiger

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astronomystudent said:
Isn't it 456 meters, not km? I am finishing the problem. It is due in an hour..
No, what you wrote in your other post is right -- 456 km and ~456000 meters. For the mass, just invert the calculation you were doing before.
 
Okay I did it for the last time I think. And the numbers are worse, but I doubled checked it.

diameter = 456927.1982 meters
mass = 3.3913079 x 10^-25 kg
volume = 4.995047837 X 10^16 m^3
density = 6.78932795 x 10^-42 kg/m^3

I mean seriously.. i have no idea what is wrong.
 
Invert the equation, I solved for M so I could isolate the x, are you saying that is wrong?
 

SpaceTiger

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astronomystudent said:
Invert the equation, I solved for M so I could isolate the x, are you saying that is wrong?
Yes, your algebra was probably wrong. Kepler's Third Law is

[tex]P^2=\frac{4\pi^2 a^3}{GM}[/tex]

Solving for M gives:

[tex]M=\frac{4\pi^2 a^3}{GP^2}[/tex]
 
ok so the mass is 2.36 x10^16 kg and the density is 4.72269734.. so can i say the density = 4.72 kg/m^3?
 
And no idea as to what it is composed of..
 

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astronomystudent said:
ok so the mass is 2.36 x10^16 kg
You're getting there, but this number is still much too small. Check to make sure your distances are in meters, your time is in seconds, and G=6.67e-11.
 
You're right. I squared instead of cubed a
the mass = 2.95x10^24 kg
and the density = 59032871.67 kg/m^3
how can i rewrite the density, and how do i figure out what that is composed of?
 

tony873004

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What it's made of will be your guess. Your options are pretty much: ice, rock, metal. It would be odd to find something of that size be pure metal, and being that it's in the Kuiper Belt, should give you a clue. But this is hypothetical, so don't rule anything out until you compare your values to the density of ice, rock and metal (iron). And it can be a combination of these things. Just take your best guess. That's all scientists do with this small amount of data.
 
I am gonna go with rock. Thank you very much for all of ya'lls help. It is greatly appreciated.
 

tony873004

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astronomystudent said:
You're right. I squared instead of cubed a
the mass = 2.95x10^24 kg
and the density = 59032871.67 kg/m^3
how can i rewrite the density, and how do i figure out what that is composed of?
You could express it in grams / cm^3. Just divide by 1000.

You got the right answers.

This is where your teacher is weird! This object is more than half as massive as Earth, while only ~1/30 Earth's diameter. That's why it is so dense. It is a hypothetical question, so anything's possible. I'd like to know what your teacher says is the correct answer for its composition.

You should try to do #4 now, even though you don't need it. You almost had it and that's a good problem to understand.
 
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Yeah, I went to class, and he told us the answer was METAL. And I asked why it was metal and he said it was much too dense to be rock or ice. Is taht true?
 

tony873004

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astronomystudent said:
Yeah, I went to class, and he told us the answer was METAL. And I asked why it was metal and he said it was much too dense to be rock or ice. Is taht true?
Yes, but it's also too dense to be metal. Ask him what kind of metal.
 

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